# How do you evaluate the integral int x"arcsec"(x^2)?

May 20, 2018

$\int x \cdot a r c \sec \left({x}^{2}\right) \cdot \mathrm{dx}$

=$\frac{1}{2} {x}^{2} \cdot a r c \sec \left({x}^{2}\right) - \frac{1}{2} \ln \left({x}^{2} + \sqrt{{x}^{4} - 1}\right) + C$

#### Explanation:

$\int x \cdot a r c \sec \left({x}^{2}\right) \cdot \mathrm{dx}$

=$\frac{1}{2} \int 2 x \cdot a r c \sec \left({x}^{2}\right) \cdot \mathrm{dx}$

After using $y = {x}^{2}$ and $2 x \cdot \mathrm{dx} = \mathrm{dy}$ transforms, this integral became

$\frac{1}{2} \int a r c \sec y \cdot \mathrm{dy}$

After using $z = a r c \sec y$, $y = \sec z$ and $\mathrm{dy} = \sec z \cdot \tan z \cdot \mathrm{dz}$ transforms, it became

$\frac{1}{2} \int z \cdot \sec z \cdot \tan z \cdot \mathrm{dz}$

=$\frac{1}{2} z \cdot \sec z - \frac{1}{2} \int \sec z \cdot \mathrm{dz}$

=$\frac{1}{2} z \cdot \sec z - \frac{1}{2} \int \frac{\sec z \cdot \left(\sec z + \tan z\right) \cdot \mathrm{dz}}{\sec z + \tan z}$

=$\frac{1}{2} z \cdot \sec z - \frac{1}{2} \ln \left(\sec z + \tan z\right) + C$

For $y = \sec z$, $\tan z$ must be equal to $\sqrt{{y}^{2} - 1}$. Thus,

$\frac{1}{2} y \cdot a r c \sec y - \frac{1}{2} \ln \left(y + \sqrt{{y}^{2} - 1}\right) + C$

=$\frac{1}{2} {x}^{2} \cdot a r c \sec \left({x}^{2}\right) - \frac{1}{2} \ln \left({x}^{2} + \sqrt{{x}^{4} - 1}\right) + C$